Problem 50 - Master Resource Book in JEE Main Physics

Question

The first line of Balmer series has wavelength \(6563 \AA\). What will be the wavelength of the first member of Lyman series? (a) \(1215.4 \mathrm{~A}\) (b) \(2500 \mathrm{~A}\) (c) \(7500 \mathrm{~A}\) (d) \(600 \mathrm{~A}\)

Step-by-Step Solution

Answer

The wavelength of the first member of the Lyman series is (a) 1215.4 Å.

Step 1: Understand the Balmer Series

The Balmer series consists of transitions where electrons fall to the second energy level \((n_2=2)\). The first line in this series is the transition from \(n_1=3\) to \(n_2=2\) with a given wavelength of \(6563 \, \AA\).

Step 2: Understand the Lyman Series

The Lyman series consists of transitions of electrons falling to the first energy level \((n_2=1)\). The first member of the Lyman series refers to the transition from \(n_1=2\) to \(n_2=1\).

Step 3: Use Rydberg Formula for Wavelength

The Rydberg formula for calculating the wavelength of light emitted is: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \] where \(R_H\) is the Rydberg constant \(1.097 \times 10^7 \, \text{m}^{-1}\).

Step 4: Calculate Wavelength for Lyman Series

Substitute \(n_2 = 1\) and \(n_1 = 2\) into the formula to find \(\lambda\) for the Lyman first line:\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]This simplifies to:\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \left( 1 - \frac{1}{4} \right) = 1.097 \times 10^7 \times \frac{3}{4} \]\[ \frac{1}{\lambda} = 8.2275 \times 10^6 \]Finally, take the reciprocal to find \(\lambda\):\[ \lambda = \frac{1}{8.2275 \times 10^6} = 1.215 \times 10^{-7} \, \text{m} = 1215 \, \AA \]

Step 5: Match with Given Answers

The calculated wavelength \(1215 \, \AA\) closely matches option (a) \(1215.4 \, \AA\). Therefore, the correct answer is (a).

Key Concepts

Lyman series

The Lyman series is a set of spectral lines that result when electrons in hydrogen atoms transition to the lowest energy level, the first energy level, indicated as \(n_2=1\). This series is most commonly observed in the ultraviolet region of the electromagnetic spectrum. Understanding the Lyman series is crucial when discussing atomic spectra as it represents the simplest cases of electronic transitions.

The key transitions in the Lyman series involve electrons falling from higher levels \(n_1\) to \(n_2=1\). To give an idea:

• The transition from \(n_1=2\) to \(n_2=1\) is the first line in this series.
• Other notable transitions happen as \(n_1\) increases to 3, 4, and beyond, falling to the base level \(n_2=1\).

Each downward jump in energy levels results in the emission of a photon. The first transition \(n_1=2\) to \(n_2=1\) produces a photon of wavelength around 1215 Å, often observed in astrophysical phenomena and laboratory experiments.

Rydberg formula

The Rydberg formula is an essential tool in spectroscopy, providing a mathematical way to calculate the wavelengths of photons emitted or absorbed when an electron transitions between energy levels in an atom. Specifically for hydrogen, the formula is written as:\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]where:

• \(\lambda\) is the wavelength of the emitted light.
• \(R_H\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)).
• \(n_1\) and \(n_2\) are the principal quantum numbers associated with the higher and lower energy levels, respectively.

In hydrogen transitions like those in the Lyman or Balmer series, this formula helps predict the spectral lines one would observe. For instance, using the Rydberg formula for the Lyman series, where \(n_2=1\) and \(n_1=2\) yields the wavelength for the series' first line. This powerful equation underscores the quantized nature of electronic orbits in atoms.

wavelength calculation

Wavelength calculation is central to understanding spectra. It relies on formulas like the Rydberg formula to compute the precise wavelengths of different spectral lines.In the context of the Lyman series, calculating the wavelength of the first line involves:

• Recognizing that you are looking for the transition \(n_1=2\) to \(n_2=1\).
• Applying the Rydberg formula where \( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \).
• Using terms from the formula: \(\frac{1}{1^2} - \frac{1}{2^2}\).

This process provides the reciprocal of the wavelength:\[\frac{1}{\lambda} = 1.097 \times 10^7 \times \left( 1 - \frac{1}{4} \right) = 8.2275 \times 10^6\]When solved, \(\lambda\) equates to approximately 1215 Å. Therefore, one needs to take the reciprocal of the calculated value to get the wavelength, reinforcing the relation between frequency, energy levels, and the nature of light in atomic transitions.